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Chapter 5: Problem 25

The steady-state diffusion flux through a metal plate is \(5.4 \times 10^{-10} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\) at a temperature of \(727^{\circ} \mathrm{C}(1000 \mathrm{~K})\) and when the concentration gradient is \(-350 \mathrm{~kg} / \mathrm{m}^{4} .\) Calculate the diffusion flux at \(1027^{\circ} \mathrm{C}(1300 \mathrm{~K})\) for the same concentration gradient and assuming an activation energy for diffusion of \(125,000 \mathrm{~J} / \mathrm{mol}\).

Short Answer

Expert verified
Question: Calculate the diffusion flux at a temperature of 1300 K, given an initial diffusion flux of \(5.4 \times 10^{-10} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\) at 1000 K, a concentration gradient of \(-350 \mathrm{~kg} / \mathrm{m}^{4}\), and an activation energy for diffusion of \(125,000 \mathrm{~J} / \mathrm{mol}\). Answer: Using the step-by-step solution, we can calculate the diffusion_flux at the final temperature, \(\varphi_2\), by following the steps and using the given values.

Step by step solution

01

Identify given values

From the problem statement, we are given: - Initial diffusion flux, \(\varphi_1 = 5.4 \times 10^{-10} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\) - Initial temperature, \(T_1 = 1000 \mathrm{~K}\) - Concentration gradient, \(\Delta c = -350 \mathrm{~kg} / \mathrm{m}^{4}\) - Final temperature, \(T_2 = 1300 \mathrm{~K}\) - Activation energy for diffusion, \(E_a = 125,000 \mathrm{~J} / \mathrm{mol}\) Our goal is to find the diffusion flux at the final temperature, \(\varphi_2\).
02

Calculate the diffusion coefficient at the initial temperature

First, we can use Fick's first law of diffusion to determine the diffusion coefficient, \(D_1\), at the initial temperature: $$ \varphi_1 = -D_1 \Delta c $$ Calculate \(D_1\): $$ D_1 = -\frac{\varphi_1}{\Delta c} $$
03

Formulate the Arrhenius equation for diffusion coefficients

The Arrhenius equation relates the diffusion coefficients at two different temperatures: $$ \frac{D_1}{D_2} = e^{\frac{-E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})} $$ Where \(R\) is the gas constant (\(8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\)).
04

Solve for the diffusion coefficient at the_final temperature, \(D_2\)

Insert all given values and the \(D_1\) into the Arrhenius equation for diffusion coefficients and solve for \(D_2\): $$ D_2 = D_1 \cdot e^{\frac{-E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})} $$
05

Use Fick's first law to determine the diffusion_flux at the final temperature, \(\varphi_2\)

Now we can use the diffusion coefficient at the final temperature, \(D_2\), along with the concentration gradient to determine the diffusion_flux at the final temperature using Fick's first law of diffusion: $$ \varphi_2 = -D_2 \Delta c $$ By following these steps, you will find the diffusion_flux at the final temperature, \(\varphi_2\).

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Most popular questions from this chapter

Carbon is allowed to diffuse through a steel plate \(15 \mathrm{~mm}\) thick. The concentrations of carbon at the two faces are \(0.65\) and \(0.30 \mathrm{~kg} \mathrm{C} / \mathrm{m}^{3}\) Fe, which are maintained constant. If the preexponential and activation energy are \(6.2 \times\) \(10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) and \(80,000 \mathrm{~J} / \mathrm{mol}\), respectively, compute the temperature at which the diffusion flux is \(1.43 \times 10^{-9} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\).

The activation energy for the diffusion of carbon in chromium is \(111,000 \mathrm{~J} / \mathrm{mol}\). Calculate the diffusion coefficient at \(1100 \mathrm{~K}\left(827^{\circ} \mathrm{C}\right)\), given that \(D\) at \(1400 \mathrm{~K}\left(1127^{\circ} \mathrm{C}\right)\) is \(6.25 \times\) \(10^{-11} \mathrm{~m}^{2} / \mathrm{s}\)

Self-diffusion involves the motion of atoms that are all of the same type; therefore, it is not subject to observation by compositional changes, as with interdiffusion. Suggest one way in which self-diffusion may be monitored.

The outer surface of a steel gear is to be hardened by increasing its carbon content. The carbon is to be supplied from an external carbon-rich atmosphere, which is maintained at an elevated temperature. A diffusion heat treatment at \(850^{\circ} \mathrm{C}(1123 \mathrm{~K})\) for \(10 \mathrm{~min}\) increases the carbon concentration to \(0.90 \mathrm{wt} \%\) at a position \(1.0 \mathrm{~mm}\) below the surface. Estimate the diffusion time required at \(650^{\circ} \mathrm{C}(923 \mathrm{~K})\) to achieve this same concentration also at a) 1.0-mm position. Assume that the surface carbon content is the same for both heat treatments, which is maintained constant. Use the diffusion data in Table \(5.2\) for C diffusion in \(\alpha\)-Fe.

Briefly explain the difference between selfdiffusion and interdiffusion.
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